Problem: Find the modular inverse of $27$, modulo $28$.

Express your answer as an integer from $0$ to $27$, inclusive.
We are looking for an integer $a$ such that $27a$ is congruent to 1 modulo 28. In other words, we want to solve  \[
27 a \equiv 1 \pmod{28}.
\]We subtract $28a$ from the left-hand side to obtain $-a\equiv 1 \pmod{28}$. This congruence is equivalent to the previous one since $28a$ is a multiple of 28. Next we multiply both sides by $-1$ to obtain $a\equiv -1\pmod{28}$. Thus $28-1=\boxed{27}$ is the modular inverse of 27 (mod 28). (Note that since $(m-1)^2=m^2-2m+1\equiv 1\pmod{m}$, we always have that $m-1$ is its own inverse modulo $m$.)